\(\int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx\) [78]

Optimal result

Integrand size = 21, antiderivative size = 137 \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {5 d^3 \sin (a+b x)}{2 b \sqrt {d \tan (a+b x)}}+\frac {d^3 \sin ^3(a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {5 d^2 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{4 b}+\frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

[Out]

5/2*d^3*sin(b*x+a)/b/(d*tan(b*x+a))^(1/2)+d^3*sin(b*x+a)^3/b/(d*tan(b*x+a))^(1/2)+5/4*d^2*csc(b*x+a)*(sin(a+1/
4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sin(2*b*x+2*a)^(1/2)*(d*tan(b*x+a))^
(1/2)/b+2/3*d*sin(b*x+a)^3*(d*tan(b*x+a))^(3/2)/b

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2674, 2678, 2681, 2653, 2720} \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=\frac {d^3 \sin ^3(a+b x)}{b \sqrt {d \tan (a+b x)}}+\frac {5 d^3 \sin (a+b x)}{2 b \sqrt {d \tan (a+b x)}}-\frac {5 d^2 \sqrt {\sin (2 a+2 b x)} \csc (a+b x) \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right ) \sqrt {d \tan (a+b x)}}{4 b}+\frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \]

[In]

Int[Sin[a + b*x]^3*(d*Tan[a + b*x])^(5/2),x]

[Out]

(5*d^3*Sin[a + b*x])/(2*b*Sqrt[d*Tan[a + b*x]]) + (d^3*Sin[a + b*x]^3)/(b*Sqrt[d*Tan[a + b*x]]) - (5*d^2*Csc[a
+ b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])/(4*b) + (2*d*Sin[a + b*x]^3*
(d*Tan[a + b*x])^(3/2))/(3*b)

Rule 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2674

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sin[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] - Dist[b^2*((m + n - 1)/(n - 1)), Int[(a*Sin[e + f*x])^m*(
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n] &&  !(GtQ[m,
1] &&  !IntegerQ[(m - 1)/2])

Rule 2678

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-b)*(a*Sin
[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] + Dist[a^2*((m + n - 1)/m), Int[(a*Sin[e + f*x])^(m - 2)*(b*
Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ
[2*m, 2*n]

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\left (3 d^2\right ) \int \sin ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx \\ & = \frac {d^3 \sin ^3(a+b x)}{b \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{2} \left (5 d^2\right ) \int \sin (a+b x) \sqrt {d \tan (a+b x)} \, dx \\ & = \frac {5 d^3 \sin (a+b x)}{2 b \sqrt {d \tan (a+b x)}}+\frac {d^3 \sin ^3(a+b x)}{b \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{4} \left (5 d^2\right ) \int \csc (a+b x) \sqrt {d \tan (a+b x)} \, dx \\ & = \frac {5 d^3 \sin (a+b x)}{2 b \sqrt {d \tan (a+b x)}}+\frac {d^3 \sin ^3(a+b x)}{b \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {\left (5 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{4 \sqrt {\sin (a+b x)}} \\ & = \frac {5 d^3 \sin (a+b x)}{2 b \sqrt {d \tan (a+b x)}}+\frac {d^3 \sin ^3(a+b x)}{b \sqrt {d \tan (a+b x)}}+\frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b}-\frac {1}{4} \left (5 d^2 \csc (a+b x) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx \\ & = \frac {5 d^3 \sin (a+b x)}{2 b \sqrt {d \tan (a+b x)}}+\frac {d^3 \sin ^3(a+b x)}{b \sqrt {d \tan (a+b x)}}-\frac {5 d^2 \csc (a+b x) \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{4 b}+\frac {2 d \sin ^3(a+b x) (d \tan (a+b x))^{3/2}}{3 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.61 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12 \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=-\frac {\csc (a+b x) \sqrt {\sec ^2(a+b x)} \left (120 \sqrt [4]{-1} \cos (2 (a+b x)) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right ),-1\right )+(22+77 \cos (2 (a+b x))+22 \cos (4 (a+b x))-\cos (6 (a+b x))) \sqrt {\sec ^2(a+b x)} \sqrt {\tan (a+b x)}\right ) (d \tan (a+b x))^{5/2}}{48 b \tan ^{\frac {3}{2}}(a+b x) \left (-1+\tan ^2(a+b x)\right )} \]

[In]

Integrate[Sin[a + b*x]^3*(d*Tan[a + b*x])^(5/2),x]

[Out]

-1/48*(Csc[a + b*x]*Sqrt[Sec[a + b*x]^2]*(120*(-1)^(1/4)*Cos[2*(a + b*x)]*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[
Tan[a + b*x]]], -1] + (22 + 77*Cos[2*(a + b*x)] + 22*Cos[4*(a + b*x)] - Cos[6*(a + b*x)])*Sqrt[Sec[a + b*x]^2]
*Sqrt[Tan[a + b*x]])*(d*Tan[a + b*x])^(5/2))/(b*Tan[a + b*x]^(3/2)*(-1 + Tan[a + b*x]^2))

Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.73 (sec) , antiderivative size = 1840, normalized size of antiderivative = 13.43

method	result	size
default	\(\text {Expression too large to display}\)	\(1840\)

[In]

int(sin(b*x+a)^3*(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/48/b*tan(b*x+a)*(-6*I*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a
))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(b*x+a)^2+6*I*EllipticPi((1+csc(
b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/
2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*cos(b*x+a)^2-6*I*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(
1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(b*x+a
)+6*I*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(
b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*cos(b*x+a)-6*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/
2),1/2+1/2*I,1/2*2^(1/2))*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x
+a))^(1/2)*cos(b*x+a)^2+72*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*
x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)^2-6*(1+csc(b*x+a)-cot(b*x+a))^(1
/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(cot(b*x+a)-csc(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),
1/2-1/2*I,1/2*2^(1/2))*cos(b*x+a)^2+8*2^(1/2)*cos(b*x+a)^4*sin(b*x+a)-6*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*
x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2+1/2*I,
1/2*2^(1/2))*cos(b*x+a)+72*(cot(b*x+a)-csc(b*x+a))^(1/2)*(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*
x+a))^(1/2)*EllipticF((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)-6*(cot(b*x+a)-csc(b*x+a))^(1/2)*
(-csc(b*x+a)+1+cot(b*x+a))^(1/2)*(1+csc(b*x+a)-cot(b*x+a))^(1/2)*EllipticPi((1+csc(b*x+a)-cot(b*x+a))^(1/2),1/
2-1/2*I,1/2*2^(1/2))*cos(b*x+a)-3*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-2*cot(b*x+a)*2^(1/2)*(-c
ot(b*x+a)*csc(b*x+a)^2*(-1+cos(b*x+a))^2)^(1/2)-2*csc(b*x+a)*2^(1/2)*(-cot(b*x+a)*csc(b*x+a)^2*(-1+cos(b*x+a))
^2)^(1/2)-2*cot(b*x+a)+2)*cos(b*x+a)^2+3*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(2*cot(b*x+a)*2^(1/
2)*(-cot(b*x+a)*csc(b*x+a)^2*(-1+cos(b*x+a))^2)^(1/2)+2*csc(b*x+a)*2^(1/2)*(-cot(b*x+a)*csc(b*x+a)^2*(-1+cos(b
*x+a))^2)^(1/2)-2*cot(b*x+a)+2)*cos(b*x+a)^2-6*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((-sin(b*
x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*x+a)^2+6*(-c
os(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)
^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*x+a)^2-52*cos(b*x+a)^2*sin(b*x+a)*2^(1/2)-3*(-cos(b*x+a)*sin(b*
x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(-2*cot(b*x+a)*2^(1/2)*(-cot(b*x+a)*csc(b*x+a)^2*(-1+cos(b*x+a))^2)^(1/2)-2*csc
(b*x+a)*2^(1/2)*(-cot(b*x+a)*csc(b*x+a)^2*(-1+cos(b*x+a))^2)^(1/2)-2*cot(b*x+a)+2)*cos(b*x+a)+3*(-cos(b*x+a)*s
in(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*ln(2*cot(b*x+a)*2^(1/2)*(-cot(b*x+a)*csc(b*x+a)^2*(-1+cos(b*x+a))^2)^(1/2)+2
*csc(b*x+a)*2^(1/2)*(-cot(b*x+a)*csc(b*x+a)^2*(-1+cos(b*x+a))^2)^(1/2)-2*cot(b*x+a)+2)*cos(b*x+a)-6*(-cos(b*x+
a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((-sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1
/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*x+a)+6*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)*arctan((sin(b*
x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))*cos(b*x+a)-16*sin(
b*x+a)*2^(1/2))*(d*tan(b*x+a))^(1/2)*d^2/(cos(b*x+a)^2-1)*2^(1/2)

Fricas [F]

\[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \sin \left (b x + a\right )^{3} \,d x } \]

[In]

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(-(d^2*cos(b*x + a)^2 - d^2)*sqrt(d*tan(b*x + a))*sin(b*x + a)*tan(b*x + a)^2, x)

Sympy [F(-1)]

Timed out. \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(sin(b*x+a)**3*(d*tan(b*x+a))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=\int { \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} \sin \left (b x + a\right )^{3} \,d x } \]

[In]

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*tan(b*x + a))^(5/2)*sin(b*x + a)^3, x)

Giac [F(-2)]

Exception generated. \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(sin(b*x+a)^3*(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0,0]ext_reduce Error: Bad Argument TypeThe choice was done assuming 0=[0,0]
ext_reduce

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(a+b x) (d \tan (a+b x))^{5/2} \, dx=\int {\sin \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2} \,d x \]

[In]

int(sin(a + b*x)^3*(d*tan(a + b*x))^(5/2),x)

[Out]

int(sin(a + b*x)^3*(d*tan(a + b*x))^(5/2), x)